Wallace’s version of the Deutsch-Wallace theorem, part 1

One might still be worried about Deutsch’s Additivity. What if it is actually necessary prove the Born rule? In this case one wouldn’t be able to use the Born rule in the Many-Worlds interpretation without committing oneself to stupid decisions, such as giving away all your money to take part in St. Petersburg’s lottery. Should one give up on the Many-Worlds interpretation then? Or start betting against the Born rule? If these thoughts are keeping you awake at night, then you need Wallace’s version of the Deutsch-Wallace theorem, that replaces Deutsch’s simplistic decision theory with a proper one that allows for bounded utilities.

Wallace’s insight was to realise that the principles of Indifference and Substitution do all the real work in Deutsch’s argument: they are already enough to imply the mod-squared amplitude part of Born’s rule. The connection of those mod-squared amplitudes with probabilities then follow from the other, decision-theoretical principles, but those are incidental, and can be replaced wholesale with a proper decision theory.

More precisely, Wallace used Indifference and Substitution to prove[1]Actually he completely changed the principles, using new ones called Erasure, Branching indifference, Microstate indifference, and Diachronic consistency, but they amount to the same thing. a theorem called Equivalence, which states that Amir must be indifferent between games that assign equal Born-rule weights to the same rewards.

It was not at all obvious to me why this should be a strong result. After all, if I say that the games \[ \ket{G} = \alpha\ket{M_0}\ket{r_0}+\beta\ket{M_1}\ket{r_1}\] and
\[ \ket{G’} = \gamma\ket{D_0}\ket{r_0}+\delta\ket{D_1}\ket{r_1}\] are equivalent if $|\alpha|^2=|\gamma|^2$ and $|\beta|^2=|\delta|^2$, it will also be true that they are equivalent if $|\alpha|=|\gamma|$ and $|\beta|=|\delta|$[2]In fact for any bijective function of the modulus of the amplitudes, so we haven’t actually learned anything about the “square” part of the Born rule, we have only learned that the phases of the amplitudes are irrelevant. Or have we?

Actually, Equivalence shows its power only when we consider sums of mod-squared amplitudes. It says, for example, that the game (taken unnormalised for clarity) \[ \ket{G} = 2\ket{M_0}\ket{r_0}+\ket{M_1}\ket{r_1}\] is equivalent to the game
\[ \ket{G’} = \ket{M_0}\ket{r_0}+\ket{M_1}\ket{r_0}+\ket{M_2}\ket{r_0}+\ket{M_3}\ket{r_0}+\ket{M_4}\ket{r_1},\] as they both assign weight $4$ to reward $r_0$ and weight $1$ to reward $r_1$. Some alternative version of Equivalence that summed the modulus of the amplitudes instead, as it would be appropriate in classical probability theory, would claim that $G$ was actually equivalent to
\[ \ket{G’^\prime} = \ket{M_0}\ket{r_0}+\ket{M_1}\ket{r_0}+\ket{M_2}\ket{r_1},\] as they both would assign weight $2$ to reward $r_0$ and weight $1$ to reward $r_1$, a decidedly non-quantum result.

Having hopefully convinced you that Equivalence is actually worthwhile, let’s proceed to prove it. The proof is actually very similar to the one presented in the previous post, so if you think it is obvious how to adapt it you can safely skip to the next post, where we’ll do the decision-theory part of the proof. Below I’ll write down the proof of Equivalence anyway just for shits and giggles.

First let’s state it properly:

Equivalence: Consider two games \[ \ket{G} = \sum_{ij}\alpha_{ij}\ket{M_i}\ket{r_j}\quad\text{and}\quad \ket{G’} = \sum_{ij}\beta_{ij}\ket{D_i}\ket{r_j}.\] If all rewards $r_j$ have the same Born-rule weight, that is, if \[\forall j\quad \sum_i|\alpha_{ij}|^2 = \sum_i|\beta_{ij}|^2,\] then $G \sim G’$.

Note that unlike in the previous post we’re not stating that these games have the same value, but rather that Amir is indifferent between them, which we represent with the $\sim$ relation. We do this because we want to eventually prove that Amir’s preferences can be represented by such a value function, so it feels a bit inelegant to start with the assumption that it exists.

Now, let’s recall Indifference and Substitution from the previous post, slightly reworded to remove reference to the values of the games:

Indifference: If two games $G$ and $G’$ differ only by the labels of the measurements, then $G \sim G’$.

And to the proof. First we show that any complex phases are irrelevant. For that, consider the game
\[ \ket{G} = \alpha e^{i\phi}\ket{M_0}\ket{r_0}+\beta e^{i\varphi}\ket{M_1}\ket{r_1}.\] By Substitution, we can replace the rewards $r_0$ and $r_1$ with the degenerate games $e^{-i\phi}\ket{D_0}\ket{r_0}$ and $e^{-i\varphi}\ket{D_1}\ket{r_1}$, and Amir must be indifferent between $G$ and the game \[ \ket{G’} = \alpha \ket{M_0}\ket{D_0}\ket{r_0}+\beta\ket{M_1}\ket{D_1}\ket{r_1}.\]Since $G’$ can be obtained from a third game \[\ket{G’^\prime} = \alpha \ket{M_0}\ket{r_0}+\beta \ket{M_1}\ket{r_1}\] via Substitution, this accumulation of measurements does not matter either, and we have Amir must be indifferent to any phases.

This allows us to restrict our attention to positive amplitudes. It does not, however, allow us to restrict our attention to amplitudes which are square roots of rational numbers, but we shall do it anyway because the argument for all real numbers is boring. Consider then two games \[ \ket{G} = \sum_{ij}\sqrt{\frac{p_{ij}}{q_{ij}}}\ket{M^j_i}\ket{r_j}\quad\text{and}\quad \ket{G’} = \sum_{ij}\sqrt{\frac{a_{ij}}{b_{ij}}}\ket{D^j_i}\ket{r_j}\] for which
\[\forall j\quad \sum_i\frac{p_{ij}}{q_{ij}} = \sum_i\frac{a_{ij}}{b_{ij}}.\] We shall show that $G \sim G’$. First focus on the reward $r_0$. We can rewrite the amplitudes of the measurement results that give $r_0$ so that they have the same denominator in both games by defining $d_0 = \prod_i q_{i0}b_{i0},$ and the integers $p_{i0}’ = d_0 p_{i0}/q_{i0}$ and $a_{i0}’ = d_0 a_{i0}/b_{i0}$, so that
\[ \frac{p’_{i0}}{d_{0}} = \frac{p_{i0}}{q_{i0}}\quad\text{and}\quad\frac{a’_{i0}}{d_{0}} = \frac{a_{i0}}{b_{i0}}.\]The parts of the games associated to reward $r_0$ are then
\[ \frac1{\sqrt{d_0}}\sum_i \sqrt{p_{i0}’}\ket{M^0_i}\ket{r_0} \quad\text{and}\quad \frac1{\sqrt{d_0}}\sum_i \sqrt{a_{i0}’}\ket{D^0_i}\ket{r_0}, \] and using again Substitution we replace the reward given for measurement results $M^0_i$ and $D^0_i$ with the trivial games \[\frac1{\sqrt{p_{i0}’}}\sum_{k=1}^{p_{i0}’}\ket{P_k}\ket{r_0}\quad\text{and}\quad \frac1{\sqrt{a_{i0}’}}\sum_{k=1}^{a_{i0}’}\ket{P_k}\ket{r_0}, \] obtaining
\[ \frac1{\sqrt{d_0}}\sum_i\sum_{k=1}^{p_{i0}’}\ket{M^0_i}\ket{P_k}\ket{r_0} \quad\text{and}\quad \frac1{\sqrt{d_0}}\sum_i \sum_{k=1}^{a_{i0}’}\ket{D^0_i}\ket{P_k}\ket{r_0}, \] which are just uniform superpositions, with $\sum_ip_{i0}’$ terms on the left hand side and $\sum_ia_{i0}’$ on the right hand side. Judicious use of Indifference and Substitution can as before erase the differences in the piles of measurements, taking them to
\[ \frac1{\sqrt{d_0}}\sum_{l=1}^{\sum_ip_{i0}’}\ket{C_l}\ket{r_0} \quad\text{and}\quad \frac1{\sqrt{d_0}}\sum_{l=1}^{\sum_ia_{i0}’}\ket{C_l}\ket{r_0}. \] Now by assumption we have that \[\sum_i\frac{p’_{i0}}{d_{0}} = \sum_i\frac{p_{i0}}{q_{i0}} = \sum_i\frac{a_{i0}}{b_{i0}} = \sum_i\frac{a’_{i0}}{d_{0}},\]so the number of terms on both sides are the same, so the $r_0$ parts of the games are equivalent. Since this same argument can be repeated for all other $r_j$, Equivalence is proven.